Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Unitary Operator on a Tensor Product

  1. Apr 7, 2010 #1
    Having a little trouble deriving a result in a book.

    If I have an operator of the form [tex]e^{\alpha A \otimes I_n}[/tex]

    Where alpha is a complex constant, A a square hermitian matrix and I the identity matrix.

    Now if I want to operator that on a tensor product, say for instance [tex]c_{n,1} |1 \rangle \otimes |n \rangle[/tex] then how would I do that?

    I firstly used the identity [tex]e^{\alpha A \otimes I_n} = e^{\alpha A \otimes I_n} [/tex] to obtain:
    [tex]e^{\alpha A \otimes I_n} {} | c_{n,1} |1 \rangle \otimes |n \rangle \rangle = e^{\alpha A} \otimes I_n {} |c_{n,1} {}|1\rangle \otimes |n \rangle \rangle = c_{n,1} e^{\alpha A}{}|1\rangle \otimes |n \rangle [/tex]

    but my book gets [tex]c_{n,1} e^{\alpha}|1 \rangle \otimes |n \rangle [/tex] with no further explanation. By the way the form of the matrix [tex] A = \begin{pmatrix} 1 & 0 & 0 \\0 & 0 & 0 \\0 & 0 & -1 \end{pmatrix} [/tex] if it helps to know it.

    Also I am aware you can represent [tex]e^{\alpha A}[/tex] in the form of an infinite series but I don't see how that helps here. In fact I tried it and I didn't know where I should cut the series off at, and it gave me coefficients of alpha rather than e^alpha. Oh and the n's are being summed over, not sure if that makes any difference.
    Last edited: Apr 7, 2010
  2. jcsd
  3. Apr 7, 2010 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    [tex]e^{A\otimes B}|\alpha\rangle\otimes|\beta\rangle=\sum_{n=0}^\infty \frac{1}{n!}(A\otimes B)^n|\alpha\rangle\otimes|\beta\rangle=\sum_{n=0}^\infty \frac{1}{n!}\underbrace{(A\otimes B)\cdots(A\otimes B)}_{n\text{ factors}}|\alpha\rangle\otimes|\beta\rangle[/tex]

    [tex]=\sum_{n=0}^\infty \frac{1}{n!}\underbrace{(A\otimes B)\cdots(A\otimes B)}_{n-1\text{ factors}}A|\alpha\rangle\otimes B|\beta\rangle=\dots=\sum_{n=0}^\infty \frac{1}{n!}A^n|\alpha\rangle\otimes B^n|\beta\rangle[/tex]

    Now what do you get when you set B=I?
  4. Apr 8, 2010 #3
    Thank you for your reply Fredrik, however I'm still stuck.

    Using your definition [tex]\sum_{n=0}^\infty \frac{1}{n!}A^n|\alpha\rangle\otimes B^n|\beta\rangle[/tex] applied to the above expression, I obtain:

    [tex]\sum_{k=0}^\infty \frac{1}{k!}(\alpha A)^k|1\rangle\otimes|n\rangle[/tex]

    I cannot quite see how to turn that into [tex] e^{\alpha}|1 \rangle \otimes |n \rangle [/tex]

    What happens to the matrix A? The vector [tex]|1\rangle[/tex] is an eigenvector of A with eigenvalue 1.

    So can I express the above as:

    [tex]\sum_{k=0}^\infty \frac{1}{k!}(\alpha A)^k|1\rangle \otimes|n\rangle = \sum_{k=0}^\infty \frac{1}{k!}(\alpha)^k A^k|1\rangle \otimes|n \rangle [/tex] and then work out [tex]A^k |1\rangle[/tex]

    However, I don't know how to work out [tex]A^k |1\rangle[/tex] ? I know that [tex]A |1\rangle = 1 |1\rangle = |1\rangle [/tex] but how can I use this to calculate the action of A^k?
  5. Apr 8, 2010 #4
    Hold on, I have solved it!

    [tex]A^k = A.A...A[/tex] k times therefore [tex]A^k |1 \rangle = |1\rangle[/tex]

    Thank you Fredrik :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook